evaluate 17(1-1/2^2)(1-1/3^2)...(1-1/17^2)

Question

please also show me the steps by steps solution and the quickest way possible to get the answer,thanks.

Answer 1

17 17

Let y = 17∏(1-1/n²) = 17∏(n-1)(n+1)/n² = 17•(1•2•3•4..16)•(3•4•5...18)/((2•3•4..17)•(2•3•4...17))

n=2 n=2

y = 17•16!•(18!/2!)/(17!•17!) = 16!•17!•(18/2)/(16!•17!) = 18/2 = 9

Answer 2

The multiplication terms are 17 * the product of (1 - 1/n^2 = (n^2-1)/n^2 = (n-1)(n+1)/n^2) for n = 2 to 17

Thus, we have 17 * (2-1)(2+1)/2^2 * (3-1)(3+1)/3^2 * (4-1)(4*1)/4^2...*(15-1)(15+1)/15^2*(16-1)(16+1)/16^2 * (17-1)(17+1)/17^2

Note how, for each of n=3 to 16, the squared term in the denominator is balanced out by the n+1 numerator in the previous term and the n-1 term in the succeeding term

For 2 and 17, we see the term once in the numerator (n-1) at n = 3 and (n+1) at n = 16

We also see the (n+1) term in the numerator for n = 17 and the n-1 term = 1 for n = 1;

Thus, we have 1 * 18/ (2 * 17)

We multiply the product times 17, eliminating the 17.

Thus, our solution is 18/2 = 9